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Electricity is the chapter that brings the flow of electric charge to life, helping you understand how circuits work and why resistance matters in everyday electronics! This chapter is one of the most interesting and scoring topics in the Class 12 Physics syllabus. Here, you'll find easy-to-understand NCERT solutions for questions 3.1 to 3.9, along with a few important additional questions from the chapter to give you extra practice. These solutions are prepared by Physics experts and are fully aligned with the latest CBSE syllabus to help you learn and revise better.
The JEE Mains 2025 April 3 shift 1 exam commenced at 9 am. The JEE Mains April 3 shift 1 exam 2025 will conclude at 12 noon.
These NCERT solutions are a great way to build your confidence and boost your exam preparation. They not only help you with board exam questions but are also super useful if you’re preparing for JEE, NEET, or other entrance exams. So, make sure to go through each solution carefully and practice the additional questions to strengthen your understanding of Current Electricity!
Download the free PDF of Class 12 Physics Chapter 3 Exercise Solutions and get exam-ready for your CBSE board! These detailed solutions cover all the important questions, helping you strengthen your understanding of Current Electricity and score high in exams.
NCERT solutions for class 12 physics chapter 3 Current Electricity: Exercise Question
Answer :
Given, the emf of battery, E = 12 V
Internal resistance of battery, r = 0.4 Ohm
Let I be the maximum current drawn from the battery.
We know, according to Ohm's law
E = Ir
I = E/r = 12/0.4 =30 A
Hence the maximum current drawn from the battery is 30 A.
Answer:
Given, The emf of the battery, E = 10 V
The internal resistance of the battery, r = 3 Ohm
Current in the circuit, I = 0.5 A
Let R be the resistance of the resistor.
Therefore, according to Ohm's law:
E = IR' = I(R + r)
10 = 0.5(R + 3)
R = 1 Ohm
Also,
V = IR (Across the resistor)
= 0.5 x 17 = 8.5 V
Hence, the terminal voltage across the resistor = 8.5 V
Answer:
Given,
temperature coefficient of filament,
α = 1.7 x 10-4 oC-1
Let T2 be the temperature of element, R2=117 Ω
(Positive alpha means that the resistance increases with temperature. Hence we can deduce that T2 will be greater than T1 )
We know,
Hence, the temperature of the element is 1027 °C.
Answer:
Hence, the resistivity of the material of wire is
Answer:
Given,
We know,
2.7 = 2.1[1 +
Hence, the temperature coefficient of silver wire is 0.0039
Answer:
For the given voltage, the two values of current will correspond to two different values of resistance which will correspond to two different temperature.
V = 230 V
Using Ohm's law:
and
Let
We know,
That is
230/2.8 = 230/3.2[1 + (
T 2 = (840.5 + 27) °C
Hence, steady temperature of the element is 867.5 °C.
3.7 Determine the current in each branch of the network shown in Fig. 3.30:
Answer:
The current in the circuit is distributed like
where I1, I2, and I3 are the different currents through shown branches.
Now, applying KVL in Loop
Also, we have
So putting it in KVL equation
Now let's apply KVL in the loop involving I1 I2 AND I3
now, the third equation of KVL
Now we have 3 equation and 3 variable, on solving we get
Now the total current
Answer:
Emf of battery, E = 8 V
Internal resistance of battery, r = 0.5 Ω
Supply Voltage, V = 120 V
The resistance of the resistor, R = 15.5 Ω
Let V' be the effective voltage in the circuit.
Now, V' = V - E
V' = 120 - 8 = 112 V
Now, current flowing in the circuit is:
I = V' / (R + r)
Now, using Ohm ’s Law:
Voltage across resistor R is v = IR
V = 7 x 15.5 = 108.5 V
Now, the voltage supplied, V = Terminal voltage of battery + V
The purpose of having a series resistor is to limit the current drawn from the supply.
Answer:
We know,
vd :drift Velocity = length of wire(l) / time taken to cover
by substituting the given values
Therefore, the time required by an electron to drift from one end of a wire to its other end is
Answer:
Given, the surface charge density of earth
Current over the entire globe = 1800 A
Radius of earth, r = 6.37 x 106 m
=
Now, charge on the earth surface,
Therefore,
Let the time taken to neutralize earth surface be t
t = 282.78 s.
Therefore, time take to neutralize the Earth's surface is 282.78 s
Answer:
Given,
There are 6 secondary cells.
Emf of each cell, E = 2 V (In series)
The internal resistance of each cell, r = 0.015 Ω (In series)
And the resistance of the resistor, R = 8.5 Ω
Let I be the current drawn in the circuit.
Hence current drawn from the supply is 1.4 A
Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V
Answer:
Given,
Emf , E = 1.9 V
Internal resistance, r =380 Ω
The maximum current that can drawn is I = E/r = 1.9/380 = 0.005 A
The motor requires a large value of current to start and hence this cell cannot be used for a motor of a car.
Answer:
We know,
R = Ρ l / A
The wires have the same resistance and also are of the same length.
Hence,
Now, mass = Density x Volume = Density x Area x length
Taking the ratio of their masses for the same length
Hence,
Therefore, for the same resistance and length, the aluminium wire is lighter.
Since aluminium wire is lighter, it is used as power cables.
5. What conclusion can you draw from the following observations on a resistor made of alloy manganin?
Current A |
Voltage V |
Current A |
Voltage V |
0.2 |
3.94 |
3.0 |
59.2 |
0.4 |
7.87 |
4.0 |
78.8 |
0.6 |
11.8 |
5.0 |
98.6 |
0.8 |
15.7 |
6.0 |
118.5 |
1.0 |
19.7 |
7.0 |
138.2 |
2.0 |
39.4 |
8.0 |
158.0 |
Answer:
The ratio of Voltage to current for the various values comes out to be nearly constant which is around 19.7.
Hence the resistor made of alloy manganin follows Ohm's law.
Answer the following questions:
Answer:
The current flowing through the conductor is constant for a steady current flow.
Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.
Answer the following questions:
Answer:
No. Ohm’s law is not universally applicable for all conducting elements.
A semiconductor diode is such an example.
Answer the following questions
8. A low voltage supply from which one needs high currents must have very low internal resistance. Why?
Answer:
Ohm's law states that: V = I x R
Hence for a low voltage V, resistance R must be very low for a high value of current.
Answer the following questions:
9. A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
Answer:
A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.
10. Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (
Answer:
(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 1022 .
11. (a)
Answer:
To get maximum effective resistance, combine them in series. The effective resistance will be nR.
11. (b)
Answer:
To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.
11.(c)
(iii) What is the ratio of the maximum to minimum resistance?
Answer:
The ratio is nR/(R/n) =
12.(a)
Answer:
We have, equivalent resistance = 11/3
Let's break this algebraically so that we can represent it in terms of 1, 2 and 3
this expression is expressed in terms of 1, 2 and 3. and hence we can make a circuit which consist only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :
12. (b)
Answer:
Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it
Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1
R = 11/5 Ω
12.(c)
Answer:
1 Ω+2 Ω+ 3 Ω= 6 Ω, so we will combine the resistance in series.
12.(d)
Answer:
Connect all three resistors in parallel.
Equivalent resistance is R = 1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)
R = 6/11 Ω
13.
(a) Determine the equivalent resistance of networks shown in
Fig. 3.31.
Answer:
It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor and two 2 ohms are also in series and we have 4 loops,
so equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is
now we have 4 such loops in series so,
Hence equivalent resistance of the circuit is 16/3 ohm.
13.
(b) Determine the equivalent resistance of networks shown in Fig. 3.31.
Answer:
It can be seen that all 5 resistors are in series, so
Equivalent Resistance = R + R + R + R + R = 5R
Hence equivalent resistance is 5R.
Answer:
First, let us find the equivalent of the infinite network,
let equivalent resistance = R'
Here from the figure, We can consider the box as a resistance of R'
Now, we can write,
equivalent resistance = R'
' =[( R')Parallel with (1)] + 1 + 1
Since resistance can never be negative we accept
, We have calculated the equivalent resistance of infinite network,
Now
Total equivalent resistance = internal resistance of battery+ equivalent resistance of the infinite network
= 0.5+1+1.73
=3.23 ohm
Hence current drawn from the 12V battery is 3.72 Ampere.
Answer:
Given
Maintained constant emf of standard cell = 1.02V, balanced point of this cell = 67.3cm
Now when the standard cell is replaced by another cell with emf =
Now as we know the relation
Hence, emf of another cell is 1.247V.
(b) What purpose does the high resistance of 600 k
Answer:
If a sufficiently high current passes through galvanometer then it can get damaged. So we limit the current by adding a high resistance of 600 k
(c) Is the balance point affected by this high resistance?
Answer:
No, the Balance point is not affected by high resistance. High resistance limits the current to galvanometer wire. The balance point is obtained by moving the joe key on the potentiometer wire and current through potentiometer wire is constant. The balance point is the point when the current through galvanometer becomes zero. The only duty of high resistance is to supply limited constant current to potentiometer wire.
Answer:
No, the method would not have worked if the driver cell of the potentiometer had an emf of 1.0V instead of 2, because when emf of the driving point is less than the other cell, there won't be any balance point in the wire.
(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
Answer:
No, the circuit would not work properly for very low order of Voltage because the balance points would be near point A and there will be more percentage error in measuring it. If we add series resistance with wire AB. It will increase the potential difference of wire AB which will lead to a decrease in percentage error.
20.
Answer:
Given,
the balance point of cell in open circuit =
value of external resistance added =
new balance point =
let the internal resistance of the cell be
Now as we know, in a potentiometer,
Hence, the internal resistance of the cell will be 1.68
Chapter 3 of Class 12 Physics NCERT Solutions is crucial for students preparing for board exams, JEE, and NEET. This chapter introduces fundamental concepts like electric current and resistance, which are not only important for scoring well in board exams but also essential for competitive exams like JEE and NEET. The solutions help build a strong foundation for these topics, and understanding them is key for advancing in physics and electrical engineering. So, mastering this chapter is vital for both exam success and future studies!
Current density
Where: is the angle between the normal to the area and direction of current
Relation between current density and electric field-
where = conductivity and = resistivity or specific resistance of the substance
Ohm’s Law
Resistivity
Where:
m is the mass, n is the number of electrons per unit volume, e is the charge of the electron and is the relaxation time
Grouping of Resistance
Series Grouping of resistance
Parallel Grouping of Resistance
The power developed
Kirchoff's Voltage Law(KVL)
Wheatstone's Bridge
Current Electricity (Chapter 3) in Class 12 Physics is important for both board exams and competitive exams like JEE Main and NEET. The topics covered in this chapter are essential, and NCERT Solutions for this chapter will help you understand them thoroughly.
Here’s a breakdown of the key topics in Chapter 3 - Current Electricity:
1. Electric Current
This part explains the concept of electric current, drift velocity, mobility, and other related concepts. All the formulas and concepts here are crucial for both the CBSE board exams and competitive exams like JEE Main and NEET.
2. Ohm's Law
Ohm's Law shows the relationship between voltage and current at a given temperature. It also covers ohmic and non-ohmic conductors, which follow or don't follow the law. This section introduces resistance, and questions based on Ohm’s Law are included in the NCERT Solutions for Class 12 Physics Chapter 3.
3. Resistance and Resistivity
You’ll learn about resistance and resistivity and how they change with temperature. The color code of resistance is also explained here. Make sure you understand these concepts, as there are practice questions based on them in the NCERT solutions.
4. Cells and Their Combinations
This section explains the concepts of cells and how they work in series and parallel combinations. You’ll definitely encounter questions on this in your exams.
5. Kirchhoff's Laws
Here, you'll study Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL), which are essential for solving electrical circuits. These laws are important for competitive exams like JEE Main and NEET.
6. Wheatstone's Bridge, Meter Bridge, and Potentiometer
You’ll learn about the Wheatstone Bridge, the Meter Bridge and how they work, including their balanced conditions. The potentiometer is also introduced here, along with its applications, like comparing EMFs of cells and finding internal resistance. These topics have high weightage in exams, and you can expect questions on them in NEET and JEE Main. Practice questions from these topics using the NCERT Solutions to get better at them.
When it comes to CBSE board exams and competitive exams like NEET and JEE Main, the solutions for NCERT Class 12 Physics Chapter 3 are super important. 7-10% of the questions in NEET and JEE Main are based on this chapter. In the 2019 CBSE board exam, there was even a 6-mark question on Current Electricity. So, using these NCERT solutions for Chapter 3 will help you score better in both your board exams and competitive exams.
Complete Coverage: These solutions cover all topics and questions in Chapter 3 - Current Electricity, making sure you understand everything clearly.
Easy-to-understand Explanations: Each solution breaks down concepts step-by-step, so even tough topics are easier to grasp.
Simple and Clear: The solutions are written in simple, clear language, which helps you understand the concepts without confusion.
Practice Questions: The exercises include practice questions that let you test yourself and check your understanding.
Perfect for Exam Prep: These solutions help you prepare well for your board exams, JEE, NEET, and any other competitive exams.
Foundation for Future Learning: The concepts in this chapter are essential for understanding advanced topics in physics and electrical engineering.
Links to Later Topics: The chapter connects to future chapters, showing how important it is as a foundation for your studies.
Free Access: You can get these solutions for free, making them easily available for everyone.
These features make the NCERT Class 12 Physics Chapter 3 Solutions a super helpful resource to help you ace your exams and build a strong understanding of physics!
Also, check
In CBSE board exam, around 8 to 10% questions can be expected from the chapter Current Electricity. Certain papers of CBSE ask around 15% questions from NCERT chapter 3 Current Electricity. To score well in the exam follow NCERT syllabus and the exercise given in the NCERT Book. To practice more problems can refer to NCERT exemplar.
If we analyse the previous year NEET, JEE Main and board papers we can see that the topics like meter bridge, potentiometer and problems using kcl and kvl and combinations of resistance and cells are frequently asked.
In JEE main, 2 to 3 questions can be expected from the chapter on Current Electricity. The topics like Potentiometer, meter bridge, KVL and KCL are important.
The concepts in the current electricity chapter of NCERT Class 12 Physics are integral parts of NEET exam. Along with NCERT, practicing with previous year papers and mock tests would be enough.
Current electricity carries 8% of weightage on an average for NEET exam.
Current electricity is the basics of electrical and electronics engineering-related branches. In these branches, analysis and designs of the circuit are important and the basic laws studied in the Class 12 Physics Chapter 3 NCERT solutions help for the same.
The main topics are-
Focus on understanding the key concepts, practising numerical problems, and solving previous year questions. Refer to NCERT solutions for detailed explanations and practice regularly.
Kirchhoff’s laws — Current Law (KCL) and Voltage Law (KVL) — help in analyzing complex electrical circuits. These laws are crucial for solving circuit problems, especially in JEE and NEET exams.
The Wheatstone Bridge and Potentiometer are important for measuring resistance and EMF in electrical circuits. They have a high weightage in competitive exams like JEE Main and NEET.
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Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
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